Assuming that the equations define $x$ and $y$ implicitly as differentiable functions $x=f(t),y=g(t)$, find the slope of the curve $x=f(t)$, $y=g(t)$ at the given value of $t$.
$$x=t^5+t,y+4t^5=4x+t^3,t=2$$

Step 1 ：Given the equations $x=t^5+t$ and $y+4t^5=4x+t^3$ at $t=2$, we are asked to find the slope of the curve at this point.

Step 2 ：We can find the slope of the curve at a given point by taking the derivative of y with respect to x. However, since we have the functions in terms of t, we can use the chain rule to find dy/dx. The chain rule states that dy/dx = (dy/dt) / (dx/dt).

Step 3 ：First, we need to find dy/dt and dx/dt. We have $x=t^5+t$ and $y=t^3+4t$.

Step 4 ：Taking the derivative of x with respect to t, we get $dx/dt=5t^4+1$.

Step 5 ：Taking the derivative of y with respect to t, we get $dy/dt=3t^2+4$.

Step 6 ：Using the chain rule, we find that the slope of the curve is given by $dy/dx=(dy/dt)/(dx/dt)=(3t^2+4)/(5t^4+1)$.

Step 7 ：Substituting $t=2$ into the equation for the slope, we find that the slope of the curve at $t=2$ is $16/81$.

Step 8 ：Final Answer: The slope of the curve $x=f(t)$, $y=g(t)$ at $t=2$ is $\boxed{{\displaystyle \frac{16}{81}}}$.